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For the lossless small-slope wave equation on a string, tension $T$ and linear mass density $\mu$ give wave speed $c=\sqrt{T/\mu }$. With transverse force $F_y=-T{y}_x$ exerted across a cut toward increasing $x$, a rightgoing displacement $y=f(x-ct)$ has velocity $v_y=-c{y}_x$. Its positive mechanical impedance is therefore $Z=F_y/v_y=T/c=\sqrt{T\mu }$. A leftgoing component has $F_y/v_y=-Z$ under the same fixed orientation. A rightgoing time-harmonic wave with peak displacement wave amplitude $|A|$ carries mean power $Z{\omega }^2|A{|}^2/2$.

For the reflection of a string wave from a freely moving endpoint mass, let $c=\sqrt{T/\mu }$, $k=\omega /c>0$ and use the complex time-harmonic wave conventionThe string pulls its endpoint toward negative $x$, so its transverse force on the attached mass is $-T{y}_x(0,t)$. Newton's second law gives the boundary condition $m{y}_{tt}(0,t)=-T{y}_x(0,t)$. Substitution yieldsHere $Z$ is the characteristic impedance of a stretched string. Since the incident and reflected waves share the same string, their mean power ratio is the squared modulus of this displacement reflection coefficient:There is no remaining fraction of the mean incident energy: the freely moving mass provides no dissipation and no onward transmission channel. It exchanges kinetic energy with the string during each cycle but has zero mean absorbed power. Indeed the energy flux of a transverse string wave satisfieswhose cycle average is zero. As checks, $m\to 0$ gives $R=+1$, the free-end reflection of a transverse string wave, while $m\to \infty$ at fixed $\omega >0$ gives $R=-1$, the fixed-end limit.

Use a time-harmonic wave with time dependence $e^{-i\omega t}$ and let the same symbols denote phasors. The wave amplitude vector obeysAt fixed material contrast, the stated low-frequency condition permits neglecting the diagonal frequency term to zeroth order. All remaining coefficient matrices are scalar multiples of the same exchange matrix, so they commute. Withthe transfer relation isSolve the second row for the outgoing left wave amplitude and substitute in the first. Since $C^2-H^2=1$, this gives the zero-frequency scattering through an elastic layer coefficientsThus the zero-frequency limit depends only on the endpoint seismic impedances. Monotonicity supports the given scale comparison but is not needed to integrate the zero-frequency equation. If $Z_a=Z_b$ in a monotone profile, seismic impedance is constant: there is no directional coupling, with only wave phase accumulation at nonzero frequency.

Use the time-harmonic wave convention $\mathrm{Re}(\psi e^{-i\omega t})$, with a positive background wavenumber $k$. The scalar Helmholtz equation for a constant-density acoustic model isWriting $\psi =e^{ikx}E$ gives the exact envelope equationThe paraxial approximation neglects $E_{xx}$ relative to $2ik{E}_x$, giving the parabolic wave equationThe choice of carrier and the direction of propagation matter: this is a forward, slowly varying envelope approximation. Sufficient scale conditions are $|E_{xx}|\ll 2k|E_x|$, transverse spectral components $|p|\ll k$, and medium/envelope variation on longitudinal scales large compared with $1/k$. With the carrier fixed at the background $k$, small $|n-1|$ makes the refractive phase vary slowly too. Large-angle propagation, appreciable backscattering, or rapid longitudinal variation violates the approximation. Acoustic models with variable mass density can have additional gradient terms, so the scalar Helmholtz equation itself is a model assumption. For the Gaussian beam in part (b), useful initial conditions are $kD\gg 1$ and $D/|F|\ll 1$, with the second condition absent for an initially uncurved beam. A narrow angular spectrum, rather than merely the label “Gaussian”, justifies the paraxial approximation.

Use the time-harmonic wave convention $\psi e^{-i\omega t}$ and write $\mathbf{z}=(y,z)$, $E(0,\mathbf{z})=E_0(\mathbf{z})$, with a deterministic incident wave envelope. The printed speed ratio is inconsistent with $k=k_{0}n$: the refractive index used below is $n=c_0/c$. Substituting $\psi =e^{i{k}_{0}x}E$ in the Helmholtz equation givesThe paraxial approximation discards ${\partial }_x^{2}E$. The usual weak-fluctuation model also linearizes $n^2-1=2\mu W+O({\mu }^2{W}^2)$, givingDropping the quadratic contrast is an additional weak-fluctuation assumption, not a consequence of small propagation angles. Although the linearized random potential generates attenuation of order ${\mu }^2$, it does not retain every effect of that order in the literal finite-correlation index: the discarded quadratic contrast can also produce a mean wave phase shift. This linearization must precede a Gaussian white noise limit: the square of ideal white noise has no ordinary pointwise meaning.

The split-step Fourier method alternates free-space diffraction, $U_h=e^{hD}$, with a random phase screen,For jointly Gaussian random fields, Gaussian phase averaging gives $\mathbb{E}{e}^{i{\varphi }_h}=e^{-\mathrm{Var}({\varphi }_h)/2}$. More generally, a product of $p$ fields and $q$ conjugate fields picks up $\exp (i{\sum }_a{ϵ}_a{\varphi }_h({\mathbf{z}}_a))$, with signs ${ϵ}_a=+1$ or $-1$. Its screen average is determined entirely by the wave phase covariance matrix. The deterministic diffraction step acts on each coordinate, with opposite signs on factors formed by complex conjugation. This is the basis of the field-moment equations.

There is an important closure qualification. A stationary Gaussian random field need not have independent longitudinal increments. For finite-correlation fluctuations, the unlinearized parabolic wave equation instead givesIn the weak-fluctuation model, the exact first-moment equation isThe last term cannot in general be replaced by a constant times $m$. An exact generic solution of the linearized model iswhere $\mathcal{T}$ denotes time ordering of the propagation operators. For the unlinearized model, add $i{k}_0{\mu }^2{W}^2/2$ inside the propagation generator. The covariance function of the medium is needed to evaluate this expression; One-point Gaussian distributions alone would not even determine the joint wave phase statistics.

For example, omit diffraction and take a longitudinal autocovariance function $R(s)=\mathbb{E}[W(s,\mathbf{z})W(0,\mathbf{z})]$. Direct Gaussian phase averaging givesEven here, unit-variance stationary Gaussian random fields with different $R$ give different answers. In the general problem the diffraction and multiplication operators do not commute, so this scalar attenuation cannot simply be multiplied by $U_x$ without an additional approximation.

The standard closed answer uses the Markov approximation for a random medium, made explicit in part ii. Let the longitudinally integrated covariance kernel beReplace the medium by longitudinal Gaussian white noise with this strength. A screen of thickness $h$ then has $\mathbb{E}[{\varphi }_h({\mathbf{z}}_a){\varphi }_h({\mathbf{z}}_b)]=k_0^2{\mu }^{2}hB({\mathbf{z}}_a-{\mathbf{z}}_b)$ and is independent of the incoming field. For the moment $M=\mathbb{E}{\prod }_a{E}_a^{({ϵ}_a)}$, expanding both steps to order $h$ givesIn particular, the coherent attenuation in a white-noise random medium and its Fresnel propagator solution areEquivalently, write $Wdx=d{\beta }_x$, where ${\beta }_x(\mathbf{z})$ is Brownian motion in $x$ with transverse covariance kernel $\mathbb{E}[d{\beta }_x(\mathbf{z})d{\beta }_x({\mathbf{z}}^{\prime })]=B(\mathbf{z}-{\mathbf{z}}^{\prime })dx$. The Stratonovich integral formulation is $dE=DEdx+i{k}_0\mu E\circ d{\beta }_x$. Its Itô integral form isThe mean of the Itô integral vanishes, independently confirming the attenuation drift. For $x>0$ in two transverse dimensions,Thus a unit plane wave has $m=e^{-\gamma x}$. For a general incident wave envelope, the Fresnel propagator supplies its spreading. The attenuation is redistribution between the coherent and diffuse wave fields, rather than wave absorption: for a field and its conjugate at the same point, the screen contribution in the second-moment equation cancels.

In time reversal acoustics, the array records the incoming signal, reverses each recorded time trace, and re-emits it through the same medium. This is phase conjugation in the frequency domain. For the time-harmonic wave convention $e^{-i\omega t}$, reversing a real time trace replaces its positive-frequency wave amplitude by its complex conjugate. The medium must remain unchanged between recording and re-emission.

Let $f(\mathbf{z})={\Psi }_0(0,\mathbf{z})$ and $G_L(\mathbf{\zeta },\mathbf{z})=G(L,\mathbf{\zeta };0,\mathbf{z})$, with the chosen source and array normalizations incorporated into the Green function. Let $a(\mathbf{\zeta })\ge 0$ be the array's aperture weight, equal to the indicator of its receiving region for an ideal uniform array. The recorded field isBy wave reciprocity, back-propagation has the same Green function with the source and receiver exchanged. Therefore the physically re-emitted, back-propagated wave amplitude isIf $Hf=R$ and $P_A$ multiplies by $a$, then ${\Psi }_B=H^T{P}_A\overline{Hf}$, where $T$ denotes the transpose without conjugation. Taking a final complex conjugate instead defines the adjoint reconstruction $\overline{{\Psi }_B}=H^{\ast }{P}_{A}Hf$. This distinction prevents an erroneous conjugation in the time reversal operator.

For a localized Gaussian beam or acoustic point source in a homogeneous medium, a finite aperture admits a limited range of angles. The focal width is of order $\lambda L/A$ when $A$ denotes the aperture diameter. In a random medium, multiple scattering creates paths with a larger angular spread. Each reversed path retraces its route, and the paths interfere constructively at the source. This can produce a larger effective aperture in time reversal and a narrower focus, even though the unreversed field has a complicated speckle pattern.

This comparison concerns a homogeneous reference medium; a deterministic heterogeneous medium can also provide useful multipath propagation. Suitable scale limits or frequency and spatial averaging can make refocusing self-averaging. Such self-averaging is not automatic for every monochromatic source and every random realization. Wave absorption, changing medium parameters, unresolved paths or poor array coverage can spoil refocusing. With complete capture of the propagating modes and a lossless unitary operator $H$, ideal adjoint reconstruction is already exact in either medium. Random scattering can improve finite-aperture wave focusing through angular diversity. See the regime-dependent analysis in Statistical stability in time reversal.

The instantaneous rightward energy flux along a string is the transverse force times the transverse velocity,For a rightgoing time-harmonic wave of peak displacement wave amplitude $|A|$, averaging over a wave period givesA leftgoing time-harmonic wave has the opposite signed energy flux. The positive ratios of reflected and transmitted power to incident power are thereforeAll explicit angular frequency factors cancel. In the ideal wave equation on a string, $T$ and the two linear mass densities are constant, so the mechanical impedances have no frequency or wavelength dependence. Consequently $\mathcal{T}$ is independent of wavelength even though the two string wavelengths generally differ.

Finally,This is conservation of energy at a lossless, massless junction. The transmitted displacement wave amplitude can exceed the incident one when $Z_R<Z_L$; the factor $Z_R/Z_L$ is essential when comparing power.

For a semi-infinite ideal string on $x\le 0$ with tension $T$ and linear mass density $\mu$, let a free transverse endpoint mass be $m$. Use a time-harmonic wave $y=\mathrm{Re}\{A(e^{ikx}+R{e}^{-ikx})e^{-i\omega t}\}$, where $k=\omega \sqrt{\mu /T}>0$. The endpoint boundary condition from Newton's second law is $m{y}_{tt}=-T{y}_x$. It gives the title displacement reflection coefficient, with characteristic impedance of a stretched string $Z=\sqrt{T\mu }$. Since numerator and denominator have equal moduli, all mean incident power is reflected. The endpoint kinetic energy varies within a cycle, but its mean rate of change is zero. A free mass changes the reflected phase, and does not by itself provide dissipation. The limits $m\to 0$ and $m\to \infty$ give $R=+1$ and $R=-1$ respectively at fixed positive angular frequency.